I'm looking for a theorem of the form

If $R$ is a nice ring and $v$ is a reasonable element in $R$ then Kr.Dim$(R[\frac{1}{v}])$ must be either Kr.Dim$(R)$ or Kr.Dim$(R)-1$.

My attempts to do this purely algebraically are not working, so I started looking into methods from algebraic geometry. I thought that Grothendieck's Vanishing Theorem might help (i.e. if dim$(X)=n$ then $H^i(X,\mathcal{F})=0$ for any sheaf of abelian groups $\mathcal{F}$ and any $i>n$) but the problem is that the converse for this theorem fails, so I can't conclude anything about dimension. Perhaps this theorem could give some sort of test for when dimension drops, but I'm hoping for a better answer.

We'll definitely need some hypotheses. For the application I have in mind we can assume $R$ is commutative and is finitely generated over some base ring (e.g. $\mathbb{Z}_{(2)}$), but we should not assume it's an integral domain. If necessary we can assume it's Noetherian and local, but I'd rather avoid this. As for $v$, it's not in the base ring and it has only a few relations with other elements in $R$, none of which are in the base ring. If we can't get the theorem above, perhaps we can figure out something to help me get closer:

Are there any conditions on $v$ such that the dimension would drop by more than 1 after inverting $v$?

One thing I know: to have any hope of dimension dropping by $1$ I need to be inverting a maximal irreducible component. I'm curious as to the algebraic condition this puts on $v$.

isa relation between the Krull dimension of a ring and that of its localizations (even one that can be used as an alternative definition of Krull dimension), but it involves localizations of a more subtle nature than just localizing at the powers of some $v\in R$. See Theorem 1 in hlombardi.free.fr/publis/KrullMathMonth.pdf . – darij grinberg Jun 21 '11 at 14:30